We have $A$ is an upper triangular matrix of order $n$ with $a_1,a_2,\cdots,a_n$ are the diagonal entries and rest element above the diagonal are anything.Then how can we prove that product of following matrix is zero.
$$(A-a_1I)(A-a_2I)\cdots(A-a_nI)=0$$,where $ 0$ is the zero matrix of order $n.$
MY TRY:The eigen values of matrix $A$ are the diagonal entries $a_1,a_2,\cdots,a_n$.We can also get eigen vector but i don't know how it help to solve my problem$?$ Thank you.
On
If $A$ is a upper triangular matrix, then $\lambda I- A$ a upper triangular matrix, hence
$\det(\lambda I-A)=(\lambda-a_1) \cdots(\lambda-a_n)$.
Cayley - Hamilton !
On
$\newcommand{\Set}[1]{\left\{ #1 \right\}}$$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$Since this particular argument is the key step in one of the proofs of the Cayley-Hamilton Theorem, let us prove it directly.
Vectors are column vectors. Let $e_{1}, e_{2}, \dots, e_{n}$ be the basis with respect to which $A$ is written, write $$ V_{i} = \Span{e_{i}, e_{i-1}, \dots, e_{1}}, $$ where $V_{n} = V$, and we set $V_{0} = \Set{0}$.
We have $$ V_{i} = \Span{e_{i}} \oplus V_{i-1}, $$ and $$ A e_{i} = a_{i} e_{i} + v_{i-1} $$ for some $v_{i-1} \in V_{i-1}$.
Since $$ (A - a_{n} I)(e_{n}) = 0, $$ and $$ (A - a_{n} I) e_{i} \in V_{i} $$ for $i < n$, we have $$ (A - a_{n} I) V_{n} \subseteq V_{n-1}. $$
By induction $$ (A - a_{1} I)(A - a_{2} I) \cdots (A - a_{n-1} I) $$ is zero on $V_{n-1}$, and thus $$ (A - a_{1} I)(A - a_{2} I) \cdots (A - a_{n} I) $$ is zero on $V = V_{n}$.
The proof I am referring to extends the underlying field to its algebraic closure (which may displease some), so that all matrices become triangularizable.
The characteristic polynomial $\chi_{A} = \det\big( X\mathrm{I}_n - A \big)$ is :
$$ \chi_{A}(X) = \prod \limits_{i=1}^{n} (X - a_i) $$
because $A$ is upper triangular.
It follows from Caley-Hamilton theorem that $\chi_{A}(A) = 0$.